Problem: Let $S_n$ denote the sum of the first $n$ terms of an arithmetic sequence with common difference 3.  If $\frac{S_{3n}}{S_n}$ is a constant that does not depend on $n,$ for all positive integers $n,$ then find the first term.
Answer: Let $a$ be the first term.  Then
\[S_n = \frac{n [2a + (n - 1) 3]}{2}\]and
\[S_{3n} = \frac{3n [2a + (3n - 1) 3]}{2},\]so
\[\frac{S_{3n}}{S_n} = \frac{\frac{3n [2a + (3n - 1) 3]}{2}}{\frac{n [2a + (n - 1) 3]}{2}} = \frac{3(2a + 9n - 3)}{2a + 3n - 3} = \frac{6a + 27n - 9}{2a + 3n - 3}.\]Let this constant be $c,$ so
\[\frac{6a + 27n - 9}{2a + 3n - 3} = c.\]Then $6a + 27n - 9 = 2ac + 3cn - 3c.$  Since both sides are equal for all $n,$ the coefficients of $n$ must be equal.  In other words, $27 = 3c,$ so $c = 9.$  then $6a - 9 = 18a - 27.$  Solving, we find $a = \boxed{\frac{3}{2}}.$